Chapter 2: Electrostatic Potential and Capacitance

Dear Class 12 Student! While Chapter 1 dealt with the vector side of electrostatics (Force and Field), this chapter introduces the incredibly powerful Scalar approach — Potential and Energy. You will learn how energy is stored in capacitors — the foundation of every modern electronic circuit. The derivations here (especially the Parallel Plate Capacitor, Dielectric Slab, and Energy Loss in Sharing) are guaranteed board exam questions. Let's master this chapter completely!

2.1 Electrostatic Potential and Potential Difference

Conservative Nature of the Electrostatic Force

The work done by the electrostatic force in moving a charge from point A to point B depends only on the initial and final positions and is completely independent of the path taken. This is the defining property of a conservative force. Because of this property, we can assign a unique scalar "potential energy" to every point in the electric field, just like gravitational potential energy.

Derivation: Potential due to a Single Point Charge

Let a source charge $+q$ be at origin O. We find the potential at point P, at a distance $r$ from O. We bring a test charge $+q_0$ from infinity to P very slowly (in equilibrium, no kinetic energy gained) along any path (say, radial).

At an intermediate distance $x$ from origin, the electrostatic (repulsive) force on the test charge is: $$\vec{F}_{elec} = \frac{1}{4\pi\epsilon_0} \frac{q \cdot q_0}{x^2} \hat{x} \quad \text{(directed away from O, i.e., outward)}$$

The external force required to move it without acceleration: $\vec{F}_{ext} = -\vec{F}_{elec}$ (directed inward).

Work done by external force from $\infty$ to $r$: $$W = \int_\infty^r \vec{F}_{ext} \cdot d\vec{x} = \int_\infty^r \left(-\frac{1}{4\pi\epsilon_0} \frac{q q_0}{x^2}\right)(-dx)$$ $$W = \frac{q q_0}{4\pi\epsilon_0} \int_\infty^r \frac{dx}{x^2}$$ $$= \frac{q q_0}{4\pi\epsilon_0}\left[-\frac{1}{x}\right]_\infty^r$$ $$= \frac{q q_0}{4\pi\epsilon_0}\left(-\frac{1}{r} + \frac{1}{\infty}\right) = \frac{q q_0}{4\pi\epsilon_0 r}$$

Since $V = \frac{W}{q_0}$, we get:

Superposition Principle for Potential

Since potential is a scalar quantity, the net potential at a point due to a system of $n$ charges is simply the algebraic sum of the individual potentials. No vector addition needed!

$$V_{net} = V_1 + V_2 + \cdots + V_n = \frac{1}{4\pi\epsilon_0}\sum_{i=1}^{n}\frac{q_i}{r_i}$$

where $r_i$ is the distance of the $i$-th charge from the point in question, and $q_i$ carries its sign.

2.1 (Continued) — Potential due to an Electric Dipole

A dipole consists of two equal and opposite charges $+q$ and $-q$ separated by a small distance $2a$, with dipole moment $\vec{p} = q \times 2a$ directed from $-q$ to $+q$.

Derivation: General Formula for Dipole Potential at Point P($r$, $\theta$)

Let point P be at distance $r$ from the centre O of the dipole. Let OP make angle $\theta$ with the dipole axis (direction of $\vec{p}$). Let $r_1$ = distance from $+q$ to P, and $r_2$ = distance from $-q$ to P.

For a short dipole ($r \gg a$), using the perpendicular dropped from $+q$ and $-q$ to line OP:

• $r_1 \approx r - a\cos\theta$ (distance from $+q$ to P is less when $\theta < 90°$)
• $r_2 \approx r + a\cos\theta$ (distance from $-q$ to P is more)

Net potential at P: $$V = \frac{1}{4\pi\epsilon_0}\left[\frac{q}{r_1} + \frac{(-q)}{r_2}\right]$$ $$= \frac{q}{4\pi\epsilon_0}\left[\frac{1}{r-a\cos\theta} - \frac{1}{r+a\cos\theta}\right]$$ $$V = \frac{q}{4\pi\epsilon_0}\cdot\frac{(r+a\cos\theta)-(r-a\cos\theta)}{r^2-a^2\cos^2\theta}$$ $$= \frac{q}{4\pi\epsilon_0}\cdot\frac{2a\cos\theta}{r^2-a^2\cos^2\theta}$$

Since $r \gg a$, we have $r^2 \gg a^2\cos^2\theta$, so we drop $a^2\cos^2\theta$. Also $q \times 2a = p$:

Special Cases of Dipole Potential

• On the Axial Line (towards $+q$, $\theta = 0°$): $\cos 0° = 1$. $$V_{axial(+)} = +\frac{p}{4\pi\epsilon_0 r^2} \quad \text{(positive)}$$
• On the Axial Line (towards $-q$, $\theta = 180°$): $\cos 180° = -1$. $$V_{axial(-)} = -\frac{p}{4\pi\epsilon_0 r^2} \quad \text{(negative)}$$
• On the Equatorial Plane ($\theta = 90°$): $\cos 90° = 0$. $$\boxed{V_{equatorial} = 0}$$ Zero potential at every point on the equatorial plane — contributions from $+q$ and $-q$ exactly cancel because both are equidistant from a point on the equatorial line.

2.2 Equipotential Surfaces

Definition: A surface (real or imaginary) on which the electric potential has the same value at every point is called an equipotential surface. Moving a charge along such a surface requires zero work since $\Delta V = 0$.

Shapes of Equipotential Surfaces for Different Charge Configurations

Relation Between Electric Field and Potential Gradient

Consider moving a unit positive charge through a small displacement $dr$ in the direction of field $\vec{E}$. The work done by the field: $dW = E\,dr$. By definition $dW = -dV$. Therefore:

$$\boxed{E = -\frac{dV}{dr}}$$

The quantity $\frac{dV}{dr}$ is called the potential gradient. SI unit: V/m (same as N/C for electric field).

Physical Meaning: Electric field points in the direction of decreasing potential. A positive charge at rest will naturally move from high potential to low potential (like water flowing downhill).

2.3 Potential Energy of a System of Charges

A. Potential Energy of Two Point Charges

The potential energy (U) of a system of charges equals the total work done by an external agent in assembling those charges from infinity to their present configuration (against electrostatic forces), assuming no kinetic energy is gained.

• Step 1: Bring charge $q_1$ from infinity to its position $\mathbf{r_1}$. No other charge is present, so work done $W_1 = 0$.
• Step 2: Bring charge $q_2$ from infinity to its position $\mathbf{r_2}$, at a separation $r_{12}$ from $q_1$. Potential at $r_2$ due to $q_1$ is $V = \frac{kq_1}{r_{12}}$. Work done: $W_2 = q_2 V = \frac{kq_1 q_2}{r_{12}}$.

B. Potential Energy of a System of Three or More Charges

For $n$ charges, calculate the potential energy for every distinct pair and add algebraically. For $n$ charges, there are $\dfrac{n(n-1)}{2}$ pairs.

For three charges $q_1, q_2, q_3$ at mutual separations $r_{12}, r_{23}, r_{13}$:

$$\boxed{\begin{aligned} U &= \frac{1}{4\pi\epsilon_0} \Big[ \frac{q_1 q_2}{r_{12}} \\ & \quad + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}} \Big] \end{aligned}}$$

2.4 Potential Energy in an External Field

When charges exist in an external electric field (produced by some other source charges not part of our system), we need to compute potential energy of interaction with that field as well.

A. Single Charge $q$ at Position $\vec{r}$ in External Field

If the external field creates a potential $V(\vec{r})$ at the position of charge $q$:

$$\boxed{U = qV(\vec{r})}$$

Example: A charge $+Q$ placed at a point where external potential is $V_0$ has PE = $QV_0$. If $V_0 > 0$ and $Q > 0$, then $U > 0$ (energy was stored).

B. System of Two Charges in an External Field

Consider two charges $q_1$ at $\vec{r_1}$ and $q_2$ at $\vec{r_2}$ in an external field that produces potentials $V(\vec{r_1})$ and $V(\vec{r_2})$ at their positions. The total PE has three contributions:

• Interaction of $q_1$ with the external field: $q_1 V(\vec{r_1})$
• Interaction of $q_2$ with the external field: $q_2 V(\vec{r_2})$
• Mutual interaction of $q_1$ and $q_2$: $\frac{kq_1 q_2}{r_{12}}$

$$\boxed{U = q_1 V(\vec{r}_1) + q_2 V(\vec{r}_2) + \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r_{12}}}$$

C. Potential Energy of a Dipole in a Uniform External Field — Derivation

A dipole $\vec{p}$ placed in a uniform external field $\vec{E}$ at angle $\theta$ with the field experiences a torque that tends to align it. The torque magnitude: $\tau = pE\sin\theta$.

Work done by external torque to rotate dipole from angle $\theta_1$ to $\theta_2$: $$W = \int_{\theta_1}^{\theta_2}\tau\,d\theta = pE\int_{\theta_1}^{\theta_2}\sin\theta\,d\theta = pE[-\cos\theta]_{\theta_1}^{\theta_2} = pE(\cos\theta_1 - \cos\theta_2)$$

Taking $\theta_1 = 90°$ as the reference position ($U = 0$) and $\theta_2 = \theta$: $$W = pE(\cos90° - \cos\theta) = pE(0 - \cos\theta) = -pE\cos\theta$$ This work equals the potential energy stored:

2.5 Electrostatics of Conductors

Electrostatic Shielding

The interior cavity of a hollow conductor is completely shielded from external electric fields.

• $E = 0$ inside the conductor material (by property 1).
• If the cavity is empty (no charge inside), there is no charge on the inner surface of the cavity either (provable by Gauss's Law + uniqueness theorem).
• Therefore, $\vec{E} = 0$ everywhere inside the cavity — regardless of how strong the external field is.

Applications:

• Sensitive electronic instruments enclosed in metallic boxes (Faraday cages) are shielded from external interference.
• Car chassis (metal body) protects passengers from lightning — it acts like a Faraday cage.
• Coaxial cables shield signal-carrying wires from external electromagnetic interference.

2.6 Dielectrics and Polarisation

Dielectrics are non-conducting substances (insulators) in which all electrons are tightly bound to their atoms — no free electrons. Examples: glass, mica, paper, water, rubber, wax.

Types of Dielectric Molecules

Polarisation of a Dielectric in an External Field

When a dielectric is placed in an external electric field $\vec{E}_0$:

• Non-polar molecules: The external field distorts the electron cloud relative to the nucleus. The centre of negative charge shifts slightly opposite to the field; the centre of positive charge shifts slightly in the direction of the field. Each molecule acquires an induced electric dipole moment along $\vec{E}_0$.
• Polar molecules: The field exerts torque on each permanent dipole, partially aligning them with $\vec{E}_0$. (Thermal agitation prevents perfect alignment — hence only partial alignment at room temperature.)

In both cases, the net effect is: the dielectric develops a macroscopic dipole moment per unit volume, called the Polarisation Vector $\vec{P}$.

2.7 Capacitors and Capacitance

A capacitor is a system of two conductors (called plates) separated by an insulating medium (vacuum, air, or a dielectric), designed to store electric charge and energy.

When charge $+Q$ is given to one plate, it induces charge $-Q$ on the other plate (connected to ground or being a conductor). This creates a potential difference $V$ between the plates.

Derivation: Capacitance of a Parallel Plate Capacitor (Board Exam Favourite — 5 marks)

A parallel plate capacitor consists of two large, flat, parallel conducting plates each of area $A$, separated by distance $d$, with $d \ll \sqrt{A}$ (edge effects negligible). Plate 1 carries $+Q$ (surface charge density $\sigma = Q/A$), Plate 2 carries $-Q$.

Step 1: Find Electric Field between the plates.
Each plate acts as an infinite plane of charge. Using Gauss's Law (pillbox Gaussian surface): $$E_{each\,plate} = \frac{\sigma}{2\epsilon_0}$$ Between the plates, fields from both plates add up (same direction): $$\boxed{E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}} \quad \text{(uniform, from + to − plate)}$$ Outside the capacitor, fields from both plates cancel: $E_{outside} = 0$.

Step 2: Find Potential Difference $V$ between the plates.
In a uniform field, potential difference = field × distance: $$V = E \times d = \frac{Qd}{\epsilon_0 A}$$

Step 3: Calculate Capacitance. $$C = \frac{Q}{V} = \frac{Q}{\dfrac{Qd}{\epsilon_0 A}}$$

Energy Stored in a Capacitor — Derivation

When a capacitor is charged from $q=0$ to $q=Q$, at an intermediate stage when charge is $q$ and potential is $v = q/C$, the work done in adding a further small charge $dq$: $$dW = v\,dq = \frac{q}{C}\,dq$$ Total work done (= energy stored) in charging from 0 to $Q$: $$U = \int_0^Q \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{q^2}{2}\bigg|_0^Q = \frac{Q^2}{2C}$$

2.8 Effect of Dielectric on Capacitance

A. Capacitor Completely Filled with Dielectric (Dielectric Constant $K$)

When the full space between the plates is filled with a dielectric of constant $K$:

• The electric field is reduced to $E = E_0/K$ (due to polarisation opposing the field).
• Potential difference becomes $V = Ed = E_0 d/K = V_0/K$ (since charge $Q$ is unchanged when battery is disconnected).
• New capacitance: $C = Q/V = Q/(V_0/K) = KC_0$.

$$\boxed{C_{dielectric} = KC_0 = K\frac{\epsilon_0 A}{d} = \frac{K\epsilon_0 A}{d} = \frac{\epsilon_0 \epsilon_r A}{d}}$$

where $\epsilon_r = K$ is the relative permittivity and $\epsilon = K\epsilon_0$ is the absolute permittivity of the dielectric.

B. Partial Dielectric Slab — Derivation (Important Board Question — 5 marks)

A dielectric slab of thickness $t$ (where $t < d$) and dielectric constant $K$ is inserted between the plates. The remaining gap $(d - t)$ is vacuum.

Step 1: Electric field in each region.
The surface charge density $\sigma = Q/A$ is unchanged. Using boundary conditions (normal D-field is continuous): $$E_0 = \frac{Q}{\epsilon_0 A} \quad \text{(vacuum region of thickness } d-t\text{)}$$ $$E = \frac{Q}{K\epsilon_0 A} = \frac{E_0}{K} \quad \text{(inside dielectric of thickness } t\text{)}$$

Step 2: Total potential difference $V$. $$V = E_0(d-t) + E \cdot t$$ $$= \frac{Q}{\epsilon_0 A}(d-t) + \frac{Q}{K\epsilon_0 A}\cdot t$$ $$= \frac{Q}{\epsilon_0 A}\left[(d-t) + \frac{t}{K}\right]$$

Step 3: Capacitance $C = Q/V$.

C. Battery Connected vs. Battery Disconnected — Comparison Table

When a dielectric slab (constant $K$) is inserted into a charged parallel plate capacitor, the outcome depends on whether the battery is connected:

2.9 Combination of Capacitors

A. Capacitors in Series — Derivation

When capacitors are connected end-to-end (in a single chain), they are in series.

Key property of series: Same charge $Q$ appears on every capacitor (charge cannot accumulate between the capacitors in the chain). The total voltage splits.

For $n$ capacitors $C_1, C_2, \ldots, C_n$ in series with total voltage $V$: $$V = V_1 + V_2 + \cdots + V_n = \frac{Q}{C_1} + \frac{Q}{C_2} + \cdots + \frac{Q}{C_n} = Q\left(\frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}\right)$$ Since $V = Q/C_{eq}$, dividing both sides by $Q$:

B. Capacitors in Parallel — Derivation

When all capacitors are connected between the same two nodes (same two endpoints), they are in parallel.

Key property of parallel: Same voltage $V$ appears across every capacitor. The total charge splits.

For $n$ capacitors $C_1, C_2, \ldots, C_n$ in parallel with voltage $V$: $$Q = Q_1 + Q_2 + \cdots + Q_n = C_1V + C_2V + \cdots + C_nV = V(C_1 + C_2 + \cdots + C_n)$$ Since $Q = C_{eq}V$, dividing by $V$:

2.10 Energy Loss in Sharing Charges Between Capacitors

When two charged capacitors are connected together (plate to plate), charge redistributes until both reach a common potential. During this process, energy is always lost (dissipated as heat, spark, or electromagnetic radiation) — even if the connecting wires are ideal (zero resistance). This energy loss is a fundamental result.

Derivation of Common Potential

Let capacitor $C_1$ have initial charge $Q_1$ (at potential $V_1 = Q_1/C_1$), and capacitor $C_2$ have initial charge $Q_2$ (at potential $V_2 = Q_2/C_2$). They are connected positive plate to positive plate and negative plate to negative plate.

Principle of Conservation of Charge: Total charge is conserved (no charge is created or destroyed).

Total charge: $Q_{total} = Q_1 + Q_2$. After connection, both capacitors reach common potential $V_{common}$. New charges: $Q_1' = C_1 V$ and $Q_2' = C_2 V$. By conservation:

$$Q_1' + Q_2' = Q_1 + Q_2 \Rightarrow (C_1 + C_2)V = Q_1 + Q_2$$

Derivation of Energy Loss

Initial total energy (before connection): $$U_i = \frac{1}{2}C_1 V_1^2 + \frac{1}{2}C_2 V_2^2$$

Final total energy (after connection, both at common potential $V$):
$$U_f = \frac{1}{2}(C_1 + C_2)V^2$$ $$= \frac{1}{2}(C_1+C_2)\left(\frac{C_1V_1+C_2V_2}{C_1+C_2}\right)^2$$ $$= \frac{(C_1V_1+C_2V_2)^2}{2(C_1+C_2)}$$

Energy lost:
$$\Delta U = U_i - U_f$$ $$= \frac{1}{2}C_1V_1^2 + \frac{1}{2}C_2V_2^2 - \frac{(C_1V_1+C_2V_2)^2}{2(C_1+C_2)}$$

After algebraic simplification (multiplying and expanding):
$$\Delta U = \frac{1}{2(C_1+C_2)}\Big[C_1(C_1+C_2)V_1^2$$ $$+ C_2(C_1+C_2)V_2^2 - (C_1V_1+C_2V_2)^2\Big]$$ $$\begin{aligned} &= \frac{1}{2(C_1+C_2)}\Big[C_1C_2V_1^2 \\ & \quad + C_1C_2V_2^2 - 2C_1C_2V_1V_2\Big] \end{aligned}$$

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